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User blog:Holomanga/This Is A Circuit Theory Blog Now (I)
I'm passionate about circuit theory for some reason. I'm not sure why, but it's just really nice. It's like calculus, but in real life with little calculus machines. Quantities At any point on the circuit, you have two associated quantities, which give you the state of the circuit: voltage and current. Voltage is electrical potential energy. Batteries use their own energy to give the electrons in the circuit this potential energy. If the voltage drops, then this energy is released somehow; usually as heat. This means that voltages around a closed loop must add up to 0. If you grab an electron, and then move it around the circuit, then it must lose as much energy as it gains. Otherwise, energy wouldn't be conserved. Current is the flow of charges; it's proportional to the number of electrons passing a point per second. If you have a higher current, then you have more electrons. This means that currents into and out of a junction must be equal. The number of electrons entering somewhere has to be the same as the number that leave; junctions can't create new electrons. Otherwise, charge wouldn't be conserved. Components To start with, I'll introduce two components: *Voltage source *Resistor A voltage source has two terminals. One of them is held at a voltage V above the other one. If electrons enter with voltage 0, then they will exit with voltage V. It will produce whatever current is necessary to achieve this. A resistor, when given a current I, drops the voltage down by V. The ratio between these two is the resitance R; they follow Ohm's law, V = IR. Now, we can find out the voltage and current for any circuit of voltage sources and resistors. Example: 1 Resistor Take this circuit, containing only a resistor and a voltage source. The diagram says 1k, but we can go one step better and say that the resistor has a resistance R, a current of IR, and a voltage drop of VR. The voltage source has a known voltage of V0 and an unknown current of I0. We want to find the values of IR and VR; the voltage and current across the resistor. So, let's solve this! Kirchoff's voltage law tells us that the voltages around a closed loop add to 0, so : V_0 - V_R = 0 Ohm's law tells us that the voltage and current across a resistor are related by : V_R = I_R R Kirchoff's current law tells us that the current flowing out of the voltage source is the same as the current flowing into the resistor, so : I_0 = I_R These are three equations in three unknowns, so we can solve them to find * V_R = V_0 * I_R = \frac{V_0}{R} * I_0 = \frac{V_0}{R} Neat! Example: 2 Resistors, in Series Now take this circuit, comprising two resistors in series. Both of these say 1k, but, again, let's do one better. These resistors have known resistances R1 and R2, and unknown voltages and currents V1, V2, I1 and I2. The power source has a known voltage V0, and an unknown current I0. Five unknowns! Let's find them out. Kirchoff's voltage law says voltages around a closed loop add to 0, so : V_0 - V_1 - V_2 = 0 Ohm's law tells us that the voltage and current across the resistors are related by : V_1 = I_1 R_1 : V_2 = I_2 R_2 And finally, Kirchoff's current law tells us that the currents flowing into and out of junctions are equal. In this case, our two junctions will be the wire between the two resistors, and the wire between the voltage source and the first resistor, each with one input and one output, so : I_0 = I_1 : I_1 = I_2 Now to solve this! Five equations in five unknowns, which is easy enough. We get * V_1 = \frac{V_0}{1 + \frac{R_2}{R_1}} * V_2 = \frac{V_0}{1 + \frac{R_1}{R_2}} * I_0 = \frac{V_0}{R_1 + R_2} * I_1 = \frac{V_0}{R_1 + R_2} * I_2 = \frac{V_0}{R_1 + R_2} Woo! That was fun. Now, let's consider the total voltage and current across the entire circuit, pretending that we've replaced the two resistors with a single resistor RT. V_0 = I_0 R_T V_0 = \frac{V_0}{R_1 + R_2} R_T R_T = R_1 + R_2 As you can see here, we can replace two resistors in series with a single resistor with its resistance being the sum of the omponent resistances, and we get a circuit with the exact same voltage and current! This applies generally; you can use this to simplify any circuit you see. Example: 2 Resistors, in Parallel Now let's do the same, but for two resistors in parallel. The voltage source has known voltage V0 and an unknown current I0, the resistors have known resistances R1 and R2, and they have unknown voltages and currents V1, V2, I1 and I2. The voltage law says : V_0 - V_1 = 0 : V_0 - V_2 = 0 And Ohm's law for each resistor says : V_1 = I_1 R_1 : V_2 = I_2 R_2 And Kirchoff's current law says that : I_0 = I_1 + I_2 This is five equations with five unknowns! Such good fortune. When you solve these you get * V_1 = V_0 * V_2 = V_0 * I_1 = \frac{V_0}{R_1} * I_2 = \frac{V_0}{R_2} * I_0 = \frac{V_0}{R_1} + \frac{V_0}{R_2} Well, that's very neat. Note that the voltage through each resistor is the same; this is because you can draw seperate closed loops through each resistor. However, the current is shared between the two resistors, inversely proportional to the resistance of each. Let's do what we did before, and pretend the circuit just has one big resistor, with resistance RT V_0 = I_0 R_T V_0 = \left( \frac{V_0}{R_1} + \frac{V_0}{R_2} \right) R_T \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} The reciprocal of the total resistance is the sum of the reciprocals of the restances. If you define the conductance S = R-1, then you get the nicer relationship S_T = S_1 + S_2 In other words, series circuits add the resistances, and parallel circuits add the conductances. This relation also applies generally, and you can use it to simplify circuits you come across too. It's pretty neat. Next Time Potential dividers! Capacitors and inductors! Alternating current! Ideal op-amps! Not all of those at the same time, though! It will be spread out over multiple blog posts! I hope you're excited, because I like circuits! Category:Blog posts